# How do you find a tangent line parallel to secant line?

Oct 5, 2014

You can find a tangent line parallel to a secant line using the Mean Value Theorem.

The Mean Value Theorem states that if you have a continuous and differentiable function, then

$f ' \left(x\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$

To use this formula, you need a function $f \left(x\right)$. I'll use $f \left(x\right) = - {x}^{3}$ as an example.

I'll also use $a = - 2$ and $b = 2$ for the interval for the secant line. This is the line that passes through the points $\left(- 2 , 8\right)$ and $\left(2 , - 8\right)$.

So, we know that the slope of this line will be $\frac{- 8 - 8}{2 - \left(- 2\right)} = - 4$.

To find the tangent lines parallel to this secant line, we will take the function's derivative, $f ' \left(x\right)$, and set it equal to $- 4$, then solve for $x$.

$- 3 {x}^{2} = - 4$

Solving this for $x$ gives us: x = ±sqrt(4/3).

So, the lines tangent to $y = - {x}^{3}$ at $x = \sqrt{\frac{4}{3}}$ and $x = - \sqrt{\frac{4}{3}}$ must be parallel to the secant line passing through $x = 2$ and $x = - 2$.